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Polynomials – Example

Example 1 Perform the indicated operation for each of the following.

(a) Add 6x5 − 10x2 + x − 45 to 13x2 − 9x + 4  

(b) Subtract 5x3 − 9x2 + x − 3 from x2 + x +1


Solution

(a) Add 6x5 − 10x2 + x − 45 to 13x2 − 9x + 4.

The first thing that we should do is actually write down the operation that we are being asked to do.

(6x5 − 10x2 + x − 45) + (13x2 − 9x + 4)

In this case the parenthesis are not required since are adding the two polynomials. They are there simply to make clear the operation that we are performing. To add two polynomials all that we do is combine like terms. This means that for each term with the same exponent we will add or subtract the coefficient of that term.

In this case this is,

(6x5 − 10x2 + x − 45) + (13x2 − 9x + 4) = 6x5 + ( − 10 + 13) x2 + (1 – 9) x – 45 + 4
       = 6x5 + 3x2  – 8 x – 41


(b) Subtract 5x3 − 9x2 + x − 3 from x2 + x + 1.

Again, let’s write down the operation we are doing here. We will also need to be very careful with the order that we write things down in. Here is the operation:

x2 + x + 1− (5x3 − 9x2 + x − 3)

This time the parentheses around the second term are absolutely required. We are subtracting the whole polynomial and the parenthesis must be there to make sure we are in fact subtracting the whole polynomial.

In doing the subtraction the first thing that we’ll do is distribute the minus sign through the parenthesis. This means that we will change the sign on every term in the second polynomial.
Note that all we are really doing here is multiplying a “ – 1” through the second polynomial using the distributive law. After distributing the minus through the parenthesis we again combine like terms.

Here is the work for this problem.

x2 + x + 1− (5x3 − 9x2 + x − 3) = x2 + x + 1− 5x3 + 9x2 – x + 3
   = − 5x3 + 10x2 + 4


Note that sometimes a term will completely drop out after combing like terms as the x did here.

This will happen on occasion so don’t get excited about it when it does happen.



Example 2 Multiply each of the following.

(a)   4x2 (x2 − 6x + 2)

(b)   (3x + 5)(x −10)

(c)    (4x2 − x)(6 − 3x)


Solution

(a)   4x2 (x2 − 6x + 2)

This one is nothing more than a quick application of the distributive law.

4x2 (x2 − 6x + 2) = 4x4 − 24x3 + 8x2

(b) (3x + 5)(x −10) This one will use the FOIL method for multiplying these two binomials.

First Terms Outer Terms Inner Terms Last Terms:



Recall that the FOIL method will only work when multiplying two binomials. If either of the polynomials isn’t a binomial then the FOIL method won’t work.

Also note that all we are really doing here is multiplying every term in the second polynomial by every term in the first polynomial. The FOIL acronym is simply a convenient way to remember this.

(c)    (4x2 − x)(6 − 3x)

Again we will just FOIL this one out.

(4x2 − x)(6 − 3x) = 24x2 −12x3 − 6x + 3x2 = −12x3 + 27x2 − 6x.



Example 3 Multiply each of the following.

(a)   (3x + 5)(3x − 5)

(b)   (2x + 6)2


Solution

(a)   (3x + 5)(3x − 5)

We can use FOIL on this one so let’s do that.

(3x + 5)(3x − 5) = 9x2 −15x +15x − 25 = 9x2 − 25

In this case the middle terms drop out.

(b)   (2x + 6)2

Now recall that 42 = (4)(4) = 16. Squaring with polynomials works the same way. So in this case we have,

(2x + 6)2 = (2x + 6)(2x + 6) = 4x2 +12x +12x + 36 = 4x2 + 24x + 36.


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